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Total moment of inertia of a two particle system

by Linus Pauling
Tags: inertia, moment, particle
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Linus Pauling
#1
Oct24-09, 04:49 PM
P: 190
1. Find the moment of inertia I_x of particle a with respect to the x axis (that is, if the x axis is the axis of rotation), the moment of inertia I_y of particle a with respect to the y axis, and the moment of inertia I_z of particle a with respect to the z axis (the axis that passes through the origin perpendicular to both the x and y axes).

Particle a is located 3r from the y axis, and particle b is located r away.




2. I = mr^2
I = SUMM(m_i*r_i^2)




3. I_a = 9mr^2
I_b = mr^2

Total I = 10mr^2

No idea what I am doing wrong
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Doc Al
#2
Oct24-09, 04:54 PM
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Quote Quote by Linus Pauling View Post
Particle a is located 3r from the y axis, and particle b is located r away.
Give the complete coordinates and the mass of each particle.
Linus Pauling
#3
Oct24-09, 04:57 PM
P: 190
Particle a:

distance from y axis = 3r
distance from x axis = r

Particle b:

distance from y axis = r
distance from x axis = -4r

It doesn't say anything explicitly about mass, and so I think we assume each particle has mass m.

Doc Al
#4
Oct24-09, 05:02 PM
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Total moment of inertia of a two particle system

Quote Quote by Linus Pauling View Post
3. I_a = 9mr^2
I_b = mr^2

Total I = 10mr^2

No idea what I am doing wrong
Looks like you've found I_y. What about I_x?
Linus Pauling
#5
Oct24-09, 05:06 PM
P: 190
I_x would be m(r)^2 + m(-4r)^2 = mr^2 + 16mr^2 = 17mr^2.

Am I simply adding that to I_y to get I_total? It asks for total I with respect to the y axis, though, why would I need I_x?
Linus Pauling
#6
Oct24-09, 05:10 PM
P: 190
Shoot, just realized I posted to wrong question. Here's what I am trying to answer:

Find the total moment of inertia I of the system of two particles shown in the diagram with respect to the y axis
Doc Al
#7
Oct24-09, 05:25 PM
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Quote Quote by Linus Pauling View Post
Here's what I am trying to answer:

Find the total moment of inertia I of the system of two particles shown in the diagram with respect to the y axis
Assuming the diagram matches your description, your answer seems correct to me. (Post the diagram, if you can.)
Linus Pauling
#8
Oct24-09, 05:26 PM
P: 190
Doc Al
#9
Oct24-09, 05:34 PM
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Looks good to me. I say your answer is correct.
Linus Pauling
#10
Oct24-09, 05:35 PM
P: 190
Ok, I requested the correct answer: 11mr^2

WTF?!
Doc Al
#11
Oct24-09, 05:37 PM
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Quote Quote by Linus Pauling View Post
Ok, I requested the correct answer: 11mr^2
Sounds bogus to me. You might want to mention this to your instructor.
Linus Pauling
#12
Oct24-09, 05:39 PM
P: 190
Ok, now:

Using the formula for kinetic energy of a moving particle K=\frac{_1}{^2}mv^2, find the kinetic energy K_a of particle a and the kinetic energy K_b of particle b.
Express your answers in terms of m, omega, and r separated by a comma.

I know I need to calculate linear speed, how do I do that? I know it's v=omega*r but I don't see how to put this together.....
Doc Al
#13
Oct24-09, 05:42 PM
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Quote Quote by Linus Pauling View Post
I know I need to calculate linear speed, how do I do that?
What's the relationship between tangential speed and angular speed for something going in a circle?
Linus Pauling
#14
Oct24-09, 05:44 PM
P: 190
Just edited before you replied: v=omega*r

But what are omega and r here?!
Linus Pauling
#15
Oct24-09, 05:49 PM
P: 190
Particle A:

Ka = 0.5m(3*omega*r)^2

Kb = 0.5m(omega*r)^2

Apparently Kb is wrong but I did it the same way as Ka.............
Linus Pauling
#16
Oct24-09, 05:58 PM
P: 190
Ok I got Kb = m(omega*r)^2

What is total kinteic eneregy?
Doc Al
#17
Oct25-09, 09:02 AM
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Quote Quote by Linus Pauling View Post
Ka = 0.5m(3*omega*r)^2

Kb = 0.5m(omega*r)^2

Apparently Kb is wrong but I did it the same way as Ka.............
These both look OK to me.

Quote Quote by Linus Pauling View Post
Ok I got Kb = m(omega*r)^2
How did you get that? (Are you posting this problem exactly as given, word for word?)

What is total kinetic energy?
Just add up the kinetic energy of each mass.


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